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CH. 20 ELECTROCHEMISTRY

CH. 20 ELECTROCHEMISTRY. Electrochemical Cells : classified as Galvanic Cells *spontaneous chem rxn that produce E Electolytic Cell. MAIN CONCEPTS Galvanic Cell Electolytic Cell -electroplating Cell Potentials Nernst Equation Balance 1 / 2 -Rxns

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CH. 20 ELECTROCHEMISTRY

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  1. CH. 20ELECTROCHEMISTRY Electrochemical Cells: classified as Galvanic Cells *spontaneous chem rxn that produce E Electolytic Cell MAIN CONCEPTS Galvanic Cell Electolytic Cell -electroplating Cell Potentials Nernst Equation Balance 1/2-Rxns -Acidic/Basic Solns Commercial Importance Galvanic Zn|MnO2 & Zn|Ag2Ocells in watches Fuel Cells in space crafts H2 - O2 Electrolytic NaOH purify metals electroplating

  2. Conversion Eelec to Echem ELECTROLYTIC CELLS conversion Echem to Eelec GALVANIC - VOLTAIC CELLS ELECTROLYTIC CELLS 1. process of electrolysis 2. pass electricity thru soln w/ E to cause nonspont redox 3. commercial importance-- NaOH purify ores electroplate GALVANIC CELLS 1. provides source electricity thru spont redox 2. batteries ZnC Alkaline Ag2O Pb Storage NiCd NiMH Fuel Cells

  3. REVIEW REDOX RXN OXIDATION REDUCTION gain of O atoms loss of O atoms loss of H gain of H lose e- gain e- LEO GER OIL RIG incr ox # decr ox # REDUCED H +1 ---> 0 oxidizing agent OXIDIZED Fe 0 ---> +2 reducing agent

  4. lose 2 e-, OX, ox # incr Reduction: Cu+2 (aq) + 2 e- ---> Cu0(s) Oxidation: Zn0 (s) ---> Zn+2(aq) + 2 e- Zn0 Cu0 SPONTANEOUS REDOX RXN gain 2 e-, RED, ox # decr Two ½-Reactions: No Electrical Work produced heat E released Zn+2 Cu+2

  5. BALANCING 1/2-REACTIONS Write balanced molecular, add # moles of spectator ions to get neutral cmpds

  6. Potassium permanganate & potassium iodide in basic soln Notice: K+1 spectator Step 1- step 2 not needed as balanced Step 3- step 4 Step 5 Step 6

  7. Step 7 Step 8 molecular

  8. ENERGY CAPTURE E released in spont REDOX rxn is captured to perform electrical work Work from electrochem cell E of water wheel depends on 1) vol water 2) PE of H2O w = (vol H2O)(E released/unit vol)

  9. e- e- Cathode (+) REDUCTION Anode (-) OXIDATION GALVANIC CELL Right is Reduction - Cathode (+) ox- + ne- ----> Red e- flows into cathode reacts w/ oxidized species forms reduced species Left is Oxidation - Anode (-) Red’ ---> ox-’ + ne-’ e- flow out of anode

  10. e- e- Volt Meter Cathode (+) Anode (-) salt bridge NO3- Na+ - + + - + - + - + - Zn0 Cu0 + - + - MAKE UP OF CELL consists of: 1. 2. Cu+2(aq) + Zn (s) -----> Cu (s) + Zn-2(aq) Salt bridge, electrolyte (Na+NO3-) allows slow mixing of ions 2 0.5-cells; anode(-) & cathode(+) Spont Rxn continuous e- flow thru external wire NO3-1 Ions flow thru soln of redox rxn @ eletrodes Cu+2 NO3-1 Zn+2 Zn ---> Zn+2 + 2 e- Cu+2 + 2 e- ---> Cu Zn(s)|Zn+2, 1M(aq)|| Cu+2, 1M (aq)|Cu(s)

  11. Anode Ox electrode Cathode Red electrode Salt Bridge Volt Meter Aqueous Solutions Phase Boundary Cathode (+) Anode (-) salt bridge NO3- Na+ + - + - + - - + + - Zn0 Cu0 + - - + Cu+2 Zn+2 Zn ---> Zn+2 + 2 e- Cu+2 + 2 e- ---> Cu “LINE” Notation of Electrochemical Cell Zn(s)|Zn+2, 1M(aq)|| Cu+2, 1M (aq)|Cu(s)

  12. Volt Meter H2 1 atm - - - - - Zn0 - - Pt 1 M H+ Voltmeter shows diff in electrical potential bet 2-1/2 cells known as “cell potential”; electromotive force (emf) Eocell = Eoox+ Eored Spont Rxn: Eocell> 0 Std H2 Electrode (SHE) reference electrode Eo = 0 V 2 H+1 (aq) + 2 e- <---> H20(g) What measures cell potential Eoox??? Eored??? E0cell = 0.76 V Cathode (+) Anode (-) Zn+2

  13. Std Electrode Potential in Water @ 25 oC Std Red. Pot. V Reduction 1/2-rxn 2.87 F2(g) + 2 e- -----> 2 F-1(aq) 0.80 Ag+1(aq) + 1 e- -----> Ag (s) 0.34 Cu+2(aq) + 2 e- -----> Cu (s) 0 2 H+1(aq) + 2 e- -----> H2(g) -0.28 Ni+2(aq) + 2 e- -----> Ni (s) -0.76 Zn+2(aq) + 2 e- -----> Zn(s) -3.05 Li+1(aq) + 1 e- -----> Li (s) Decrease Reduction

  14. EX: Build galvanic cell --- Ag & Zn What is reduced? Cathode? What is oxidized? Anode? What is the cell potential, emf? Highest 0.080 V Ag+1(aq) + 1 e- -----> Ag (s) Lowest -0.76 Zn(s) -----> Zn+2(aq) + 2 e- Eocell = Eoox,Zn+ Eored,Ag = +0.76 V + 0.80 V = +1.56 V Write the “line” notation Zn(s)|Zn+2, 1M(aq)|| Ag+1, 1M(aq)|Ag(s)

  15. e- - + e- Cathode (-) REDUCTION Anode (+) OXIDATION ELECTROLYTIC CELL drive a nonspontaneous reaction Eocell< 0 Left is Oxidation - Anode (+) Red ----> ox- + ne- Right is Reduction - Cathode (-) ox-’ + ne-’ = RED’ “e- pump” e- being push by external power source

  16. e- e- DC source Anode (+) Cathode (-) - + + - - + - + + - Cl20 Na0 + - + - Driven by outside source E (nonspont) ELECTROLYSIS MAKE UP OF CELL consists of: 1. 2. 2 electrodes (in molten salt) driven by DC source; e- pump Na+ ions gain e- @ cathode; reduce Cl- ions lose e- @ anode; oxidized Na+ Cl- 2 Cl- ---> Cl2 + 2 e- Na+ + e- ---> Na

  17. Electrical E (V) needed to drive NaCl rxn Oxidize Reduce 2 Cl- (aq) ---> Cl2 (g) + 2 e- Eox = -1.36 Na+ (aq) + e- ---> Na(s) Ered = -2.71 V Ecell = -4.07 V Is rxn spontaneous as written? Write balanced spont. rxn. Fe+2(aq) -----> Fe(s) + Fe+3(aq) -0.44 V -0.77 V 2 e- + Fe+2(aq) -----> Fe(s) Eo = 2( ) Fe+2(aq) -----> Fe+3(aq) + 1 e- Eo = Fe+2 + 2 Fe+2(aq) -----> Fe(s) + 2 Fe+3 (aq) 3 Fe+2(aq) -----> Fe(s) + 2 Fe+3 (aq) Eo = -1.21 V Fe(s) + 2 Fe+3(aq) -----> 3 Fe+2 (aq) Eo = +1.21 V

  18. SET #1 Using STD reduction potentials, which rxns are spontaneous? 1) I2(s) + 5 Cu+2(aq) + 6 H2O(l) ------> 2 IO3-(aq) + 5 Cu(s) + 12 H+(aq) 2) Hg2+2(aq) + 2 I-(aq) ------> 2 Hg(l) + I2(s) 3) H2SO3(aq) + 2 Mn(s) + 4 H+(aq) ------> S(s) + 2 Mn+2(aq) + 3 H2O(l) 1) OX: I2(s) + 6 H2O(l) ------> 2 IO3-(aq) + 12 H+(aq) + 10 e- RED: 5[Cu+2(aq) + 2 e- ------> Cu(s)] EOX = -1.195 ERED = 0.337 I2(s) + 5 Cu+2(aq) + 6 H2O(l) ------> 2 IO3-(aq) + 5 Cu(s) + 12 H+(aq) Ecell = 0.337 + (-1.195) = -0.858 V NONSPONT

  19. EOX = -0.536 ERED = 0.789 2) OX: 2 I-(aq) ------> I2(s) + 2 e- RED: Hg2+2(aq) + 2 e- ------> 2 Hg(l) 2 I-(aq) + Hg+2(aq) ------> I2(aq) + Hg(l) Ecell = 0.789 + (-0.536) = 0.253 V SPONT EOX = 1.18 ERED = 0.45 3) OX: 2 [Mn(s)------>Mn+2(aq) + 2 e-] RED: H2SO3(aq) + 4 H+(aq) + 4 e- ------> S(s) + 3 H2O(l) H2SO3(aq) + 2 Mn(s) + 4 H+(aq) ------> S(s) + 2 Mn+2(aq) + 3 H2O(l) Ecell = 0.45 + 1.18 = 1.63 V SPONT

  20. EMF & G Change in Free E is measure of spontaneity @ constant T & P relationship bet emf & G is: Go = -nFEo n = # e- pushed F: Faraday; 1 F = 96,500 C/mol = 96,500 J/V-mol K relationship Go = -RT LnK R=8.314 J/mol-K

  21. At 25oC, find standard Go and K for: 4 Ag(s) + O2(g) + 4 H+(aq) ------> 4 Ag+(aq) + 2 H2O(l) Need values for Eo, Go, & K Use eqn Go = -RT LnK Ered = +1.23 V Eox = -0.80 V Ecell = +0.43 V STEP1: RED: O2(g) + 4 H+(aq) + 4 e- ---> 2 H2O(l) OX: 4 Ag(s) ---> 4 Ag+(aq) + 4 e- STEP 2: Go = -nFE = -(4 e-)(96,500 J/V-mol)(+0.43 V) = -170,000 J/mol What can be deduced from this? Go very large, so very favored, expect K to be very large also

  22. STEP 3: equilib constant K Go = -RT LnK ===> Ln K = Go/-(RT) K = e69 = 9.3*1029 Notice example pg. 864 part b rxn written 1/2 of original All values remain the same, even though half of quantities K is 1/2, why????

  23. e- e- DC source + + + + + Ag0 Ag0 + + ELECTROLYTIC CELL ELECTROPLATING Anode (+) plating material Cathode (-) object to be plated Ag+ Ag+ NO3-

  24. ELECTRICAL WORK wmax = n * F * E J = (mol) * (C/mol)*(J/C) -G = wmax wmax = nFE Coulumb (C) amt of charge that passes thru a pt w/ current 1 ampere @ 1 sec 1 C = 1 A•s 1 F = 96,500 C/mol e-

  25. EX. How many grams Ag deposited from AgNO3 soln by current 1.5 Ah over 2 hrs. Steps 1. Rxn Ag deposited, gain e-, red Ag+1 (aq) + 1 e- ---> Ag (s) 2. Relationship 1 mol Ag ~ 1 mol e- = 96,500 C 3. Current & time, find C 1.5 Ah*[2*(3600 s/h)] = 10,800 A•s = 10,800 C 10,800 C*(1 mol e-/96,500 C)*(1 mol Ag/1 mol e-) *(107.8 g Ag/1 mol Ag) = 12.1 g Ag

  26. amt moles e- lost/gain in redox rxn QUANTITATIVE Michael Faraday 1st to describe extent of current used to chem change @ electrodes Amt Change - related to amt electricity passed Ag+ (aq) + e- ---> Ag (s) 1 mol Ag = 107.9 g, know 1 mol e- passed Faraday (F) amt of electricity supplied to deliver 1 mol of e-; “a mol of e-”

  27. Coulumb (C) amt of charge that passes thru a pt w/ current 1 ampere @ 1 sec 1 C = 1 A•s 1 F = 96,500 C = 1 mol e- EX. How many grams Cu deposited from CuSO4 soln by current 1.5 A over 2hrs. Steps 1. Rxn Cu deposited, gain e-, red Cu+2 (aq) + 2 e- ---> Cu (s) 2. Relationship 1 mol Cu ~ 2 mol e- = 2 F 3. Current & time, find C 1.5 A*[2*(3600 s/hr)] = 11,000 A•s = 11,000 C 11,000 C*(2 mol Cu/2 mol e-)*(1 F/96,500 C)*(63.6 g/1 mol) = 3.5 g Cu

  28. CELL POTENTIAL emf 1 V = 1 J/C Ecell = Eox + Ered Znox = -0.76 = +0.76 V Cured = +0.34 V Zn more diff to reduce Ecell = 0.76 + 0.34 = 1.10 V EX. What is the rxn and Ecell from the following: 1) Cr+3 (aq) + 3 e- -----> Cr (s) 2) MnO2 (s) + 4 H+ (aq) + 2 e- ----> Mn+2 (aq) + 2 H2O (l) STEPS 1. E1 = -0.74 V E2 = +1.28 V Red2> Red1 #1 must be oxidized, reverse 2. Rewrite, balance e-, & sum 2 [Cr(s) -----> Cr+3(aq) + 3 e-] 3[MnO2(s) + 4 H+(aq) + 2 e- ----> Mn+2(aq) + 2 H2O(l)] 3 MnO2(s) + 12 H+(aq) + 2 Cr(s) ----> 3 Mn+2(aq) + 2 Cr+3(aq) + 6 H2O(l) Ecell = 1.28 + 0.74 = 2.02 V

  29. Zn|Cu Cell move 2 e- G = (2 mol e-)(96,500 C/mol)(1.10 J/C) = @ 25oC 1 V = 1J/C (2.303RT/F) is const = 0.0592 J/C E = (0.0592/n) Log Kc G = -2.303RT Log Kc Combine Eqns nFE = 2.303 Log Kc E = (2.303RT/nF) Log Kc

  30. QUANTITATIVE, NERST EQN aA + bB ----> cC + dD @ 25oC (2.303RT/F) = 0.0592/n R: gas constant, 8.314 T: temp, K n: # e- transferred F: Faraday constant, 96,500 Q: rxn quotient Zn|Cu Cell Eo = 1.10 V n = 2 Application Nernst 1) w/ varying concentrations 2) Ksp, solubility product constant 3) pH

  31. Find the standard potential (Eo) for the rxn: Cd(s)+ Cu+2(aq) <----> Cd+2(aq)+ Cu(s) [Cu+2] = 0.80 [Cd+2] = 0.20 1/2-rxns Anode (ox) Cd(s) <----> Cd+2(aq) + 2 e- Cathode (red) Cu+2(aq)+ 2 e- <----> Cu(s) Eo 0.40 V 0.34 V Cd+2(aq)+ Cu(s) <----> Cu+2(aq)+ Cd(s) Eo = 0.40 + 0.34 = 0.74 V Nernst Eqn: E = 0.74 - 0.0592/2*Log[0.20]/[0.80] = 0.74 - 0.0296*(-0.602) = 0.75 V

  32. Concen Cell: system consists of 2 half-cells; cell #1: strip Fe metal in 1.5 M Fe+3 soln cell #2: strip Fe metal in 0.003 M Fe+3 soln What is the emf? Anode: Fe(s) --- Fe+2(aq) + 2 e- Eo = +0.44 Cathode: Fe+2(aq) + 2 e- --- Fe(s) Eo = -0.44 Fe+2(aq, conc) ---- Fe+2(aq,dilute) Eocell = 0.44 + (-0.44) = 0.0 V Nernst Eqn E = Eocell – (0.0592 V/n) Log([Fe+2]dil/[Fe+2]conc) E = 0.0 – (0.0592 V/2) Log([0.003]/[1.5]) = -(0.0296 V) Log(0.002) = -(0.0296 V)(-2.70) = 0.080 V

  33. Concentration cell with 2 Zn(s)-Zn+2(aq) half-cells. 1st half-cell [Zn+2]= 1.35 M 2nd cell [Zn+2] = 3.73*10-4 M a) which half-cell is anode? b) What is emf? 2nd Cell; more dilute concentration Oxidize Zn(s) ---> Zn+2(aq) + 2 e- Eox = 0.763 V Reduced Zn+2 (aq) +2 e- ----> Zn(s) Ered = -0.763V Eo = 0.00 V Nernst Eqn E = Eocell – (0.0592 V/n) Log([Zn+2]dil/[Zn+2]conc) Ecell = 0.00 - (0.0592/2) Log([3.73*10-4]/[1.35]) = -(0.0296)Log(2.76*10-4) = -0.0296(-3.56) Ecell = 0.105 V

  34. Ox: Anode 2 Cl- (l) ---> Cl2 (g) + 2 e- Red: Cathode 2 Na+ (l) + 2 e- ---> Na (l) emf - Electromotive Forces (V) Cell Rxn 2 Na+ (l) + 2 Cl- (l) ---> Cl2 (g) + Na (l) Ecell = Eox(Cl-) + Ered (H2O) = (-1.36) + (-.083) + -2.19 V EX. Explain why CuCl2 produces Cu (s) & Cl2(g). Min emf Soln: CuCl2(s) ----> Cu (s) & Cl2(g)

  35. Reduce 2 H2O(l) + 2 e- ---> H2(g) + 2 OH- (aq) Ered = -0.83 V cathode Cu(aq) + 2 e- ---> Cu(s) Ered = +0.34 V H2 produced @ cathode Oxidize 2 Cl- (aq) ---> Cl2 (g) + 2 e- Eox = -1.36 V anode 2 H2O(l) ---> O2(g) + 4 H+(aq) + 4 e- Eox = -1.23 V Ecell = (0.34) + (-1.36) = -1.02 V

  36. Half Reactions Ox - Anode 2 Cl- (l) ---> Cl2 (g) + 2 e- Red -Cathode 2 Na+ (l) + 2 e- ---> Na (l) Cell Rxn 2 Na+ (l) + 2 Cl- (l) ---> Cl2 (g) + Na (l) Electro- Cathode: e- forced unto (-) Anode: e- withdrawn (+) AQUEOUS SOLN more complex due to ability H2O to RED & OX Possible red-ox solvent & ions solute; whether the solute anion or H2O, or the solute cation or H2O

  37. Oxidize 2 Cl- (aq) ---> Cl2 (g) + 2 e- Eox = -1.36 V 2 H2O(l) ---> O2(g) + 4 H+(aq) + 4 e- Eox = -1.23 V Overcharge?? O2 formation is high to permit Cl- oxid than H2O “BRINES” produce H2 & Cl2 Reduce 2 H2O(l) + 2 e- ---> H2(g) + 2 OH- (aq) Ered = -0.83 V Na+ (aq) + e- ---> Na(s) Ered = -2.71 V H2 produced @ cathode Acidic Soln 2 H+ (aq) + 2 e- ---> H2(g) not major dil. aq soln

  38. Pb Storage Battery - car 12 V Consists of: 6 cells (2 V) anode: Pb Cathode: PbO2 H2SO4 (aq) Discharge Anode Pb(s) + SO4-2 (aq) -----> PbSO4(s) + 2 e- Cathode PbO2(s) + SO4-2(aq) + 4 H+(aq) + 2 e- ----> PbSO4(s) + 2 H2O(l)] Pb(s) + PbO2(s) + 4 H+(aq) + 2 SO4-2(aq) ----> 2 PbSO4(s) + 2 H2O(l) Anode: E = 0.356 V Cathode: E = 1.685 V Ecell = 0.356 + 1.685 = 2.041 V

  39. + - Reactants: Pb; PbO2 PbO2 Pb H2SO4 DH2SO4 = 1.8 Charge D = 1.25 - 1.30 Recharge D < 1.20

  40. Find the equilibrium constant for the rxn @ 25oC of Ni(s) & Ag+1 (aq) Anode: Ag 0.80 Cathode: Ni 0.25 n = 2 Ecell = 0.80 + 0.25 = 1.05 V Log K = (2 * 1.05 V)/0.0592 = 35.473 K = 1035.473 = 3*1035

  41. Find Ecell and Go @ 25oC for n = 2 & K = 5.0*10-6 Go = -RT Ln K = -(8.314 J/mol•K)*(298.15 K) Ln(5.0*10-6) = -(2478.8)*(-12.2) = 30200 J/mol Ecell = -Go/nF Ecell = -(30200 J/mol)/[(2 mol e-)*(96,500 C/mol e-)]*(V/JC-1) = -0.16 V

  42. Ex. MnO4-2 is stable in strong basic soln. (frames 6 & 7 basic example) In acidic soln, reacts to form permanganate and MnO2(s). Wrtie balanced overall rxn from the two half-rxns.

  43. Ex. Write balance eqn of CuS in 3M HNO3. Cu(s) + H+(aq) + NO3-1(aq) -----> Cu+2(aq) + S(s) + NO(g) + H2O(l)

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